Uniform circular motion

Steady speed, yet accelerating

Tie a stone to a string and whirl it round your head at a steady speed. Is the stone accelerating?

Many people say no, because the speed does not change. But the direction of the velocity changes all the time. Velocity is a vector, so a change of direction is a change of velocity. So the stone is accelerating. In this lesson we find how large that acceleration is, and which way it points.

A body that goes round a circle at a steady speed is in uniform circular motion. A stone on a string, a car on a round bend and a point on a spinning wheel are all examples.

Counting the turns

• The period $T$ is the time for one full turn.
• The frequency $f = \dfrac{1}{T}$ is the number of turns each second, in hertz (Hz).
• The angular velocity $\omega$ (the Greek letter omega) is the angle turned each second, in radians per second. One full turn is $2\pi$ radians, so

In one turn the body travels the circumference, $2\pi r$, in the time $T$. So its speed is $v = \dfrac{2\pi r}{T}$, which gives

All the points of a spinning wheel have the same $\omega$. The points farther from the centre move faster, because they go round a bigger circle in the same time.

Which way is the acceleration?

The velocity always points along the tangent, at right angles to the radius. As the body goes round, the velocity arrow turns with it. The change in velocity, over a short time, points towards the centre of the circle. So the acceleration points to the centre too. It is called the centripetal acceleration, which means "centre-seeking".

Watch the two arrows go round.

How big is it?

In a short time $\Delta t$, the body turns through a small angle $\Delta\theta$. The two radius arrows, each of length $r$, make that angle. So do the two velocity arrows, each of length $v$. The two thin triangles have the same shape, so

Divide both sides by $\Delta t$. When $\Delta t$ is very small, $\dfrac{|\Delta\vec{r}|}{\Delta t}$ becomes the speed $v$, and $\dfrac{|\Delta\vec{v}|}{\Delta t}$ becomes the acceleration $a$. So $\dfrac{a}{v} = \dfrac{v}{r}$, which gives

Worked example. A stone on a string goes round a circle of radius $0.5\ \text{m}$, twice every second.

1. Period: $T = 0.5\ \text{s}$. Angular velocity: $\omega = \dfrac{2\pi}{0.5\ \text{s}} = 12.6\ \text{rad/s}$.
2. Speed: $v = \omega r = 12.6\ \text{rad/s} \times 0.5\ \text{m} = 6.28\ \text{m/s}$.
3. Acceleration: $a = \omega^2 r = (12.6\ \text{rad/s})^2 \times 0.5\ \text{m} = 79\ \text{m/s}^2$, about $8$ times $g$. The string must pull hard to give it.

Worked example. A car goes round a bend of radius $80\ \text{m}$ at $20\ \text{m/s}$. Its acceleration is $\dfrac{(20\ \text{m/s})^2}{80\ \text{m}} = 5.0\ \text{m/s}^2$, towards the centre of the bend. The friction of the road on the tyres provides it.

Watch out. Three common mistakes.

• The acceleration is not a constant vector. Its size stays the same, but its direction turns all the time. So $v = u + at$ and the other equations of constant acceleration do not apply here.
• Nothing pulls the stone outwards. If the string breaks, the stone flies off along the tangent, in a straight line, not straight out from the centre.
• Doubling the speed on the same circle makes the acceleration four times larger, because $a$ goes with $v^2$.

Practise with new numbers

Try it in Python

The program works out the stone of the first worked example.

Remember these

• $\omega = \dfrac{2\pi}{T} = 2\pi f$, and $v = \omega r$.
• The velocity is along the tangent. The acceleration points to the centre.
• $a = \dfrac{v^2}{r} = \omega^2 r$. Its size is steady, but its direction keeps turning.

Check yourself

Try each one on your own first. If you are stuck, open a hint. When you have an answer, check it with Show the answer.

1. Is a body in uniform circular motion accelerating? Explain.

2. A wheel of radius $0.4\ \text{m}$ turns $3$ times each second. Find its angular velocity and the speed of a point on the rim.

3. A car goes round the same bend at twice the speed. How does its acceleration change?

4. A stone whirled on a string breaks free. Which way does it go?